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1.1 Introduction
Today we live in a predominantly electrical world. Electrical technology is a driving force in the changes that are occurring in every engineering discipline. For example, surveying is now done using lasers and electronic range finders. Circuit analysis is the foundation for electrical technology. An indepth knowledge of circuit analysis provides an understanding of such things as cause and effect, feedback and control and, stability and oscillations. Moreover, the critical importance is the fact that the concepts of electrical circuit can also be applied to economic and social systems. Thus, the applications and ramifications of circuit analysis are immense. In this chapter, we shall introduce some of the basic quantities that will be used throughout the text. An electric circuit or electric network is an interconnection of electrical elements linked together in a closed path so that an electric current may continuously flow. Alternatively, an electric circuit is essentially a pipe-line that facilitates the transfer of charge from one point to another.
1.2 Current, voltage, power and energy
The most elementary quantity in the analysis of electric circuits is the electric charge. Our interest in electric charge is centered around its motion results in an energy transfer. Charge is the intrinsic property of matter responsible for electrical phenomena. The quantity of charge can be expressed in terms of the charge on one electron. which is
1 602 10 ^19 coulombs. Thus, 1 coulomb is the charge on 6 24 1018 electrons. The
current flows through a specified area and is defined by the electric charge passing through that area per unit time. Thus we define as the charge expressed in coulombs. Charge is the quantity of electricity responsible for electric phenomena.
2 Network Theory
The time rate of change constitutes an electric current. Mathemetically, this relation is expressed as
or () =
(^)
The unit of current is ampere(A); an ampere is 1 coulomb per second. Current is the time rate of flow of electric charge past a given point. The basic variables in electric circuits are current and voltage. If a current flows into terminal of the element shown in Fig. 1.1, then a voltage or potential difference exists between the two terminals and. Nor- mally, we say that a voltage exists across the element.
Figure 1.1 Voltage across an element
The voltage across an element is the work done in moving a positive charge of 1 coulomb from first terminal through the element to second terminal. The unit of voltage is volt, V or Joules per coulomb. We have defined voltage in Joules per coulomb as the energy required to move a positive charge of 1 coulomb through an element. If we assume that we are dealing with a differential amount of charge and energy,
then =
Multiplying both the sides of equation (1.3) by the current in the element gives
which is the time rate of change of energy or power measured in Joules per second or watts ( ). could be either positive or negative. Hence it is imperative to give sign convention for power. If we use the signs as shown in Fig. 1.2., the current flows out of the terminal indicated by , which shows the positive sign for the voltage. In this case, the element is said to provide energy to the charge as it moves through. Power is then provided by the element.
Figure 1.2 An element with the current leaving from the terminal with a positive voltage sign
Conversely, power absorbed by an element is = , when is entering through the positive voltage terminal.
4 | Network Theory
1.3.1 Passive Circuit Elements
An element is said to be passive if the total energy delivered to it from the rest of the circuit is either zero or positive. Then for a passive element, with the current flowing into the positive (+) terminal as shown in Fig. 1.4 this means that
w =
∫^ t
−∞
vi dt ≥ 0
Examples of passive elements are resistors, capacitors and inductors.
1.3.1.A Resistors
Figure 1.5 Symbol for a resistor R
Resistance is the physical property of an ele- ment or device that impedes the flow of cur- rent; it is represented by the symbol R. Resistance of a wire element is calculated us- ing the relation:
R = ρl A
where A is the cross-sectional area, ρ the resistivity, and l the length of the wire. The practical unit of resistance is ohm and represented by the symbol Ω. An element is said to have a resistance of 1 ohm, if it permits 1A of current to flow through it when 1V is impressed across its terminals. Ohm’s law, which is related to voltage and current, was published in 1827 as v = Ri (1.6)
or R = v i where v is the potential across the resistive element, i the current through it, and R the resistance of the element. The power absorbed by a resistor is given by
p = vi = v
( (^) v R
v^2 R
Alternatively, p = vi = (iR)i = i^2 R (1.8) Hence, the power is a nonlinear function of current i through the resistor or of the voltage v across it. The equation for energy absorbed by or delivered to a resistor is w =
∫ (^) t
−∞
pdτ =
∫ (^) t
−∞
i^2 R dτ (1.9)
Since i^2 is always positive, the energy is always positive and the resistor is a passive element.
Circuit Concepts and Network Simplification Techniques 5
1.3.1.B Inductors
Whenever a time-changing current is passed through a coil or wire, the voltage across it is proportional to the rate of change of current through the coil. This proportional relationship may be expressed by the equation
Where is the constant of proportionality known as induc- tance and is measured in Henrys (H). Remember and are both funtions of time. Let us assume that the coil shown in Fig. 1.6 has turns and the core material has a high permeability so that the magnetic fluk is connected within the area . The changing flux creates an induced voltage in each turn equal to the derivative of the flux , so the total voltage across turns is
Figure 1.6 Model of the inductor
=
Since the total flux is proportional to current in the coil, we have
= (1.12)
Where is the constant of proportionality. Substituting equation (1.12) into equa- tion(1.11), we get
=
The power in an inductor is
The energy stored in the inductor is
(^)
(^) ( )
( )
^2 Joules (1.13)
Note that when = ( ) = 0. Also note that () 0 for all () so the inductor is a passive element. The inductor does not generate energy, but only stores energy.
Circuit Concepts and Network Simplification Techniques | 7
1.3.2 Active Circuit Elements (Energy Sources)
An active two-terminal element that supplies energy to a circuit is a source of energy. An ideal voltage source is a circuit element that maintains a prescribed voltage across the terminals regardless of the current flowing in those terminals. Similarly, an ideal current source is a circuit element that maintains a prescribed current through its terminals regardless of the voltage across those terminals. These circuit elements do not exist as practical devices, they are only idealized models of actual voltage and current sources. Ideal voltage and current sources can be further described as either independent sources or dependent sources. An independent source establishes a voltage or current in a circuit without relying on voltages or currents elsewhere in the circuit. The value of the voltage or current supplied is specified by the value of the independent source alone. In contrast, a dependent source establishes a voltage or current whose value depends on the value of the voltage or current elsewhere in the circuit. We cannot specify the value of a dependent source, unless you know the value of the voltage or current on which it depends. The circuit symbols for ideal independent sources are shown in Fig. 1.8.(a) and (b). Note that a circle is used to represent an independent source. The circuit symbols for dependent sources are shown in Fig. 1.8.(c), (d), (e) and (f). A diamond symbol is used to represent a dependent source.
Figure 1.8 (a) An ideal independent voltage source (b) An ideal independent current source (c) voltage controlled voltage source (d) current controlled voltage source (e) voltage controlled current source (f) current controlled current source
8 | Network Theory
1.4 Unilateral and bilateral networks
A Unilateral network is one whose properties or characteristics change with the direction. An example of unilateral network is the semiconductor diode, which conducts only in one direction. A bilateral network is one whose properties or characteristics are same in either direc- tion. For example, a transmission line is a bilateral network, because it can be made to perform the function equally well in either direction.
1.5 Network simplification techniques
In this section, we shall give the formula for reducing the networks consisting of resistors connected in series or parallel.
1.5.1 Resistors in Series
When a number of resistors are connected in series, the equivalent resistance of the com- bination is given by
R = R 1 + R 2 + · · · + Rn (1.17) Thus the total resistance is the algebraic sum of individual resistances.
Figure 1.9 Resistors in series
1.5.2 Resistors in Parallel
When a number of resistors are connected in parallel as shown in Fig. 1.10, then the equivalent resistance of the combination is computed as follows: 1 R
R 1
R 2
Rn
Thus, the reciprocal of a equivalent resistance of a parallel combination is the sum of the reciprocal of the individual resistances. Reciprocal of resistance is conductance and denoted by G. Consequently the equivalent conductance,
G = G 1 + G 2 + · · · + Gn
Figure 1.10 Resistors in parallel
10 Network Theory
Since 6Ω and 8Ω are connected in parallel, voltage across 8Ω = 12 volts. Therefore, the current through 8Ω (between A and B)
= 1 5 A
Total current in the circuit = 2 + 15 = 3 5 A Current in the 4Ω branch = 3.5 A
Current through 8Ω (betwen C and D) = 3 5
= 2.5 A
Therefore, current through 20Ω = 3 5 2 5 = 1A Total resistance of the circuit = 4 +
6 8 6 + 8
8 20 8 + 20 = 13143Ω Therefore applied voltage, = 3 5 13 143 (∵ = ) = 46 Volts
EXAMPLE 1. Find the value of in the circuit shown in Fig. 1.13.
Figure 1.
SOLUTION Voltage across 5Ω = 2 5 5 = 125 volts Hence the voltage across the parallel circuit = 25 12.5 = 12.5 volts
Current through 20Ω = 1 or^ 2
= 0625A
Circuit Concepts and Network Simplification Techniques 11
Therefore, current through = 3 = 1 2 = 2 5 0 625 0 625 = 125 Amps Hence R =
1.6 Kirchhoff’s laws
In the preceeding section, we have seen how simple resistive networks can be solved for current, resistance, potential etc using the concept of Ohm’s law. But as the network becomes complex, application of Ohm’s law for solving the networks becomes tedious and hence time consuming. For solving such complex net- works, we make use of Kirchhoff’s laws. Gustav Kirchhoff (1824-1887), an eminent German physi- cist, did a considerable amount of work on the principles governing the behaviour of eletric cir- cuits. He gave his findings in a set of two laws: (i) current law and (ii) voltage law, which together are known as Kirchhoff’s laws. Before proceeding to the statement of these two laws let us familar- ize ourselves with the following definitions encoun- tered very often in the world of electrical circuits:
Figure 1.14 A simple resistive network for difining various circuit terminologies
(i) Node: A node of a network is an equi-potential surface at which two or more circuit elements are joined. Referring to Fig. 1.14, we find that A,B,C and D qualify as nodes in respect of the above definition.
(ii) Junction: A junction is that point in a network, where three or more circuit elements are joined. In Fig. 1.14, we find that B and D are the junctions.
(iii) Branch: A branch is that part of a network which lies between two junction points. In Fig. 1.14, BAD,BCD and BD qualify as branches.
(iv) Loop: A loop is any closed path of a network. Thus, in Fig. 1.14, ABDA,BCDB and ABCDA are the loops.
(v) Mesh: A mesh is the most elementary form of a loop and cannot be further divided into other loops. In Fig. 1.14, ABDA and BCDB are the examples of mesh. Once ABDA and BCDB are taken as meshes, the loop ABCDA does not qualify as a mesh, because it contains loops ABDA and BCDB.
Circuit Concepts and Network Simplification Techniques 13
Consider the circuit shown in Fig. 1.16, where the voltage for each element is identified with its sign. The ideal wire used for connecting the components has zero resistance, and thus the voltage across it is equal to zero. The sum of voltages around the loop incorporating 6 3 4 and 5 is
6 3 + 4 + 5 = 0
The sum of voltages around a loop is equal to zero. A circuit loop is a conservative system, meaning that the work required to move a unit charge around any loop is zero. However, it is important to note that not all electrical systems are conservative. Ex- ample of a nonconservative system is a radio wave broadcasting system.
EXAMPLE 1. Consider the circuit shown in Fig. 1.17. Find each branch current and voltage across
each branch when 1 = 8Ω 2 = 10 volts 3 = 2A and 3 = 1Ω. Also find 2.
Figure 1.
SOLUTION
Applying KCL (Kirchhoff’s Current Law) at node A, we get
1 = 2 + 3
and using Ohm’s law for 3 , we get
3 =^ 3 3 = 1(2) =^ 2V
Applying KVL (Kirchhoff’s Voltage Law) for the loop EACDE, we get
10 + 1 + 3 = 0 1 = 10 3 = 8V
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Ohm’s law for R 1 is v 1 = i 1 R 1 ⇒ i 1 = v 1 R 1
= 1A
Hence, i 2 = i 1 − i 3 = 1 − 2 = −1A From the circuit, v 2 = R 2 i 2 ⇒ R 2 = v 2 i 2
EXAMPLE 1. Referring to Fig. 1.18, find the follow- ing: (a) ix if iy = 2A and iz = 0A (b) iy if ix = 2A and iz = 2iy (c) iz if ix = iy = iz
SOLUTION^ Figure 1. Applying KCL at node A, we get
5 + iy + iz = ix + 3 (a) ix = 2 + iy + iz = 2 + 2 + 0 = 4A
(b) iy = 3 + ix − 5 − iz = −2 + 2 − 2 iy
⇒ iy = 0A
(c) This situation is not possible, since ix and iz are in opposite directions. The only possibility is iz = 0, and this cannot be allowed, as KCL will not be satisfied (5 = 3).
EXAMPLE 1. Refer the Fig. 1.19.
(a) Calculate vy if iz = −3A
(b) What voltage would you need to replace 5 V source to obtain vy = −6 V if iz = 0.5A?
16 Network Theory
Applying KVL clockwise to the loop CBAC, we get
1 6 1 + 12 = 0 1 = 12 6 1 = 12 6(3) = 6volts
EXAMPLE 1. Use Ohm’s law and Kirchhoff’s law to evaluate (a) , (b) , (c) and (d) the power
provided by the dependent source in Fig 1.21.
Figure 1.
SOLUTION (a) Applying KVL, (Referring Fig. 1.21 (a)) we get
2 + + 8 = 0 = 6V
Figure 1.21(a)
Circuit Concepts and Network Simplification Techniques 17
(b) Applying KCL at node a, we get
+ 4^ +^
+ 4( 6)
24 1 5 = 4 = 29 5A
(c) Applying KCL at node b, we get
6
= 1 + 29 5
6 = 23A
(d) The power supplied by the dependent current source = 8 (4 ) = 8 4 6 = 192W
EXAMPLE 1. Find the current 2 and voltage for the circuit shown in Fig. 1.22.
Figure 1.
SOLUTION
From the network shown in Fig. 1.22, 2 = 6 The two parallel resistors may be reduced to
3 6 3 + 6
Hence, the total series resistance around the loop is
= 2 + + 4 = 8Ω
Circuit Concepts and Network Simplification Techniques 19
EXAMPLE 1. A wheatstone bridge ABCD is arranged as follows: AB = 10Ω, BC = 30Ω, CD = 15Ω
and DA = 20Ω. A 2V battery of internal resistance 2Ω is connected between points A and C with A being positive. A galvanometer of resistance 40Ω is connected between B and D. Find the magnitude and direction of the galvanometer current.
SOLUTION
Applying KVL clockwise to the loop ABDA, we get 10 + 40 20 = 0 10 20 + 40 = 0 (1.23) Applying KVL clockwise to the loop BCDB, we get 30( ) 15( + ) 40 = 0 30 15 85 = 0 (1.24) Finally, applying KVL clockwise to the loop ADCA, we get 20 + 15( + ) + 2( + ) 2 = 0 2 + 37 + 15 = 2 (1.25) Putting equations (1.23),(1.24) and (1.25) in matrix form, we get 10 20 40 30 15 85 2 37 15
Using Cramer’s rule, we find that = 0.01 A (Flows from B to D)
20 Network Theory
1.7 Multiple current source networks
Let us now learn how to reduce a network having multiple current sources and a number of resistors in parallel. Consider the circuit shown in Fig. 1.24. We have assumed that the upper node is () volts positive with respect to the lower node. Applying " to upper node yields
1 () 2 () 3 () + 4 () 5 () 6 () = 0 1 () 3 () + 4 () 6 () = 2 () + 5 () (1.26) () = 2 () + 5 () (1.27)
Figure 1.24 Multiple current source network
Figure 1.25 Equivalent circuit
where () = 1 () 3 () + 4 () 6 () is the algebraic sum of all current sources present in the multiple source network shown in Fig. 1.24. As a consequence of equation (1.27), the network of Fig. 1.24 is effectively reduced to that shown in Fig. 1.25. Using Ohm’s law, the currents on the right side of equation (1.27) can be expressed in terms of the voltage and individual resistance so that KCL equation reduces to
() =
()
Thus, we can reduce a multiple current source network into a network having only one current source.
1.8 Source transformations
Source transformation is a procedure which transforms one source into another while retaining the terminal characteristics of the original source. Source transformation is based on the concept of equivalence. An equivalent circuit is one whose terminal characteristics remain identical to those of the original circuit. The term equivalence as applied to circuits means an identical effect at the terminals, but not within the equivalent circuits themselves.
